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I hope people will indulge me a newbie question on notation.

For reference, the textbook in question is Contemporary Abstract Algebra by Joseph Gallian (8th edition).

I was reading the chapter on extension fields, which defines the Fundamental Theorem of Field Theory as follows: "Let F be a field and let f(x) be a nonconstant polynomial in F[x]. Then there is an extension field E of F in which f(x) has a zero." It then gives the following example:

Let $f(x)=x^2 + 1 \in Q[x]$. Then, viewing $f(x)$ as an element of $E[x] = (Q[x]/\langle x^2 + 1\rangle)[x]$, we have $$ f(x + \langle x^2 + 1 \rangle) = (x + \langle x^2 + 1 \rangle)^2 + 1 = x^2 + \langle x^2 + 1 \rangle + 1 = x^2 + 1 + \langle x^2 + 1 \rangle = 0 + \langle x^2 + 1 \rangle $$

I have to admit I'm slightly baffled by this example and his notation. So I understand that $Q[x]$ is a polynomial ring with rational coefficients, that $G/H$ is a factor/quotient ring (or a factor group in group theory), and that $\langle a \rangle$ is an ideal generated by $a$ (or a cyclic group generated by $a$ in group theory), but I'm having trouble understanding what this actually means when put together. Can someone help me understand this example? I'm having particular difficulty understanding what $E[x] = (Q[x]/\langle x^2 + 1\rangle)[x]$ means. What does this actually consist of (i.e. what are its elements)?

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    $\begingroup$ They should at the very least have said $(\Bbb Q[t]/\langle t^2+1\rangle)[x]$ and say that $f(t+\langle t^2+1\rangle)=0$. Using $x$ for two different things is not a good thing to do. $\endgroup$
    – Arthur
    Dec 3, 2016 at 22:12
  • $\begingroup$ @Arthur I agree, the way they wrote it is very confusing. Dumb question - $Q[t]/ \langle t^2 + 1\rangle$ is still a factor ring, right? $\endgroup$
    – EJoshuaS - Stand with Ukraine
    Dec 3, 2016 at 22:45
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    $\begingroup$ Is the variable name clash the only source of your confusion, i.e. if we instead write $\, E = \Bbb Q[t]/(t^2+1)\,$ then does that clear up the confusion? $\endgroup$
    – Bill Dubuque
    Dec 3, 2016 at 22:53

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I must say the notations are rather ambiguous. I would use another indeterminate, say $t$ and set $E=\mathbf Q[t]/(t^2+1)$. It is a field extension of $\mathbf Q$, and in this field, the polynomial $f(x)=x^2+1$ has a root, and this root is $t$ (or $-t$), since $$f(t)=t^2+1\equiv 0\mod t^2+1.$$

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  • $\begingroup$ Is there a difference between $Q[t]/ \langle t^2 + 1\rangle$ and $Q[t]/(t^2 + 1)$? I've seen both notations used. Can both (a) and $\langle a \langle$ be used to denote an ideal generated by $a$? $\endgroup$
    – EJoshuaS - Stand with Ukraine
    Dec 3, 2016 at 22:46
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    $\begingroup$ @EJoshuaS: The notations usually have the same meaning (both $(a)$ and $\langle a\rangle$ mean the ideal generated by $a$), though you may occasionally encounter authors who give one a different meaning in some special context. In my experience $(a)$ is the more common notation. $\endgroup$
    – Eric Wofsey
    Dec 3, 2016 at 22:51
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    $\begingroup$ @EJoshuaS The notation $\,(a,b)\,$ for ideals better emphasizes the important analogy with gcds (I don't ever recall seeing gcds written in angle bracket form). It's also consistent with $\!\pmod{a,b}$ notation (just omit the mod for more ideal notation!). The analogies are clarified when one studies divisor theory (modernizations of Kronecker's alternative to Dedekind's ideal theory). $\endgroup$
    – Bill Dubuque
    Dec 3, 2016 at 23:02

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